Frogger--poj2253

http://poj.org/problem?id=2253

题意：The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.
这句话很重要，题中求得是青蛙一次至少跳多远才能成功到达对方所在的地方；即求所有路径中每条路径中的最大距离的最小值（求一条1~2的路径使得路径上的最大边权最小）

floyd中更新距离改成更新最小的最大边权；

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define INF 0xfffffff
#define N 1100
using namespace std;
int n;
double maps[N][N];

struct node
{
    int x,y;
}a[N];
void Floyd()
{
    int i,j,k;
    for(k=1;k<=n;k++)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                maps[i][j]=min(maps[i][j],max(maps[i][k],maps[k][j]));
            }
        }
    }
}

int main()
{
    int i,j,k,x,y,t=1;

    while(scanf("%d",&n),n)
    {
        memset(a,0,sizeof(a));

        k=1;

        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);

            for(j=1;j<=k;j++)
            {
                double d=sqrt((x-a[j].x)*(x-a[j].x)+(y-a[j].y)*(y-a[j].y));

                maps[i][j]=maps[j][i]=d;
            }

            a[k].x=x,a[k].y=y,k++;
        }
        for(i=1;i<=n;i++)

            maps[i][i]=0;

        Floyd();

        printf("Scenario #%d
Frog Distance = %.3f

",t++,maps[1][2]);
    }
    return 0;
}

 用Dij算法就相当与prim；

dist[i] 代表i点之前的那个点到i的最短距离；求最短中的最长；

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define INF 0xfffffff
#define N 1100
using namespace std;
int n,vis[N];
double maps[N][N],dist[N],ans;
struct node
{
    int x,y;
}a[N];
void Dij()
{
    int i,j,index;
    dist[1]=0;
    for(i=1;i<=n;i++)
    {
        double Min=INF;
        for(j=1;j<=n;j++)
        {
            if(vis[j]==0&&Min>=dist[j])
                Min=dist[j],index=j;
        }
        vis[index]=1;
        if(ans<dist[index]&&dist[index]!=INF)
            ans=dist[index];
        if(index==2)
            return ;
        for(j=1;j<=n;j++)
        {
            if(vis[j]==0)
                dist[j]=min(dist[j],maps[index][j]);
        }
    }
}
int main()
{
    int i,j,k,x,y,t=1;
    while(scanf("%d",&n),n)
    {
        memset(vis,0, sizeof(vis));
        memset(a,0,sizeof(a));
        for(i=0;i<=n;i++)
            dist[i]=INF;
        k=1;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            for(j=1;j<=k;j++)
            {
                double d=sqrt((x-a[j].x)*(x-a[j].x)+(y-a[j].y)*(y-a[j].y));
                maps[i][j]=maps[j][i]=d;
            }
            a[k].x=x,a[k].y=y,k++;
        }
        for(i=1;i<=n;i++)
            maps[i][i]=0;
        ans=0;
        Dij();
        printf("Scenario #%d
Frog Distance = %.3f

",t++,ans);
    }
    return 0;
}